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### Course: Integral Calculus>Unit 5

Lesson 14: Function as a geometric series

# Power series of arctan(2x)

We can represent arctan(2x) with a power series by representing its derivative as a power series and then integrating that series. You have to admit this is pretty neat.

## Want to join the conversation?

• Wouldn't it be easier just to look at the derivative of arctan(2x) as a geometric series with the first term 2 and the ratio -4x^2? We would just write a few terms, then integrate and get to the same result. It seems easier and faster.
• Going through the geometric series route, we'd have to constrain x in such a way that our common ratio falls between -1 and 1 (Remember that a geometric series converges only when |common ratio| < 1)

Going through the Maclaurin Series route, on the other hand does not require us to constrain x in a similar way and hence, gives a more general, less constrained solution.
• Can someone explain why integrating is necessary? I think I missed that.
• we integrated so that we can walk back from the derivative that we expanded. remember, we had to first take the derivative of arctan( ) so that we could expand an equation. we need to go and account for the fact that we took the derivative of the function in our expansion so we take the anti-derivative (integral) of our f(x) function to represent the original (arctan) function.
• Wait, is a power series and a Maclaurin series the same thing?
• Almost. The difference is that Maclaurin series is "centered" at zero. Remember each single term in a power series has the form: a[sub n]*(x-c)^n In a Maclaurin series c=0, and a[sub n] equal to the (nth derivative of f)/n!
• What theorem states we can do this specifically? I think that should have been included in the video just for clarification.
• If you refer to the integration of power series, it essentially follows from the fact that a power series converges uniformly to a continuous function on, say, compact subsets of its interval of convergence.

Suppose `ƒ(x) = ∑ c(n)(x - a)ⁿ` is a power series about the point `a` with radius of convergence `R > 0`, i.e., the series converges on `(a - R, a + R)`. Then for any `0 < r < R`, the series converges uniformly to a continuous function on `[a - r, a + r]`. Since a uniformly convergent series of integrable functions is itself integrable and may be integrated term by term on such intervals, the result follows.
• could he just have used (1)/(1-x) and used a geometric series instead of 1/(1+x)?
• Yes, that would have been a valid path to take. He would have had to define `f(x) = 2·g(-4x²)`, and the end result would have been the same.

I guess he didn't do it in order to give another example of the power series.
• Hold on, what am I missing here?

Is it just a given in this video that the series expansion of the derivative of f(x) is the derivative of the series expansion of F(X) (its integral)?

Is that supposed to be obvious to me? It seems a bit glossed over, but perhaps I'm missing some key here.
• What is the best method for choosing an appropriate g(x) that is close enough to the original f(x)? This problem seems obvious, but I'm not sure I would have come up with that so readily without Sal's help. Just practice?
• In this example, Sal only uses the first 4 non-zero terms to approximate the function. That's close enough to him and to me. If you want to be more accurate, you can use more terms but it's, in some cases, quite nasty. The more term you use, the more accurate it would be
• when we substitute 4x^2 in g(x), why are we not taking the derivative again since using the chain rule, derivative of(1+4x^2), ie 8x also has to be multiplied to the derivative of g(x)?
• at when he says that one can interchange the two functions by replacing x with 4x^2, my question is why is this valid when the derivatives will have been different?

for instance if h = (1+x^2 ) ^2 than h' = 4x(1 + x^2), but if i were to let g = (1+x)^2 than g' = 2(1+x) . by replacing x^2 for x on g' != h' . so my question is why is this valid